9-1.Fluid Mechanics
hard

A small spherical ball of radius $0.1 \,mm$ and density $10^{4} \,kg m ^{-3}$ falls freely under gravity through a a distance $h$ before entering a tank of water. If after entering the water the velocity of ball does not change and it continue to fall with same constant velocity inside water, then the value of $h$ wil be $m$. (Given $g =10 \,ms ^{-2}$, viscosity of water $=1.0 \times 10^{-5} \,N - sm ^{-2}$ )

A

$10$

B

$9$

C

$30$

D

$20$

(JEE MAIN-2022)

Solution

Speed after falling through height $h$ Should be equal to terminal velocity

$\sqrt{2 gh }=\frac{2}{9} \frac{ r ^{2}( d -\rho) g }{\eta}$

$\sqrt{2 gh }=\frac{2}{9} \frac{10^{-8}(10000-1000) \times 10}{10^{-5}}$

$=\frac{2}{9} \times 10^{-8} \frac{9 \times 10^{4}}{10^{-5}}=20$

$2 \times 10 \times h =400$

$h =20 \,m$

Standard 11
Physics

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